Q. 94.2( 44 Votes )

# If the 9th term of an A.P. is zero then show that the 29^{th} term is twice the 19^{th} term.

Answer :

Now, By using n^{th} term of an A.P. formula

t_{n} = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

Given: t_{9} = 0

⇒ t_{9} = a + (9 – 1)d

⇒ 0 = a + 8d

⇒ a = – 8d

To Show: t_{29} = 2× t_{19}

Now,

⇒ t_{29} = a + (29 – 1)d

⇒ t_{29} = a + 28d

⇒ t_{29} = – 8d + 28d = 20 d (since, a = – 8d )

⇒ t_{29} = 20 d

⇒ t_{29} = 2 × 10 d ….(1)

Also,

⇒ t_{19} = a + (19 – 1)d

⇒ t_{19} = a + 18d

⇒ t_{19} = – 8d + 18d = 10 d (since, a = – 8d )

⇒ t_{19} = 10 d …..(2)

From eq. (1) and eq. (2) we get,

t_{29} = 2× t_{19}

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